SQL Server 2019

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select @@version;

(No column name)

Microsoft SQL Server 2019 (RTM-CU16) (KB5011644) - 15.0.4223.1 (X64) 
 Apr 11 2022 16:24:07 
 Copyright (C) 2019 Microsoft Corporation
 Express Edition (64-bit) on Windows Server 2019 Standard 10.0 <X64> (Build 17763: ) (Hypervisor)

DECLARE @info NVARCHAR(500)='[{"id": 1, "data": "One"}, {"id": 2, "data": "Two"}]';

PRINT @info;

SET @info = JSON_MODIFY(@info, 'append $', JSON_QUERY('{"id": 3, "data": "Three"}'))
SET @info = JSON_MODIFY(@info, 'append $', JSON_QUERY('{"id": 4, "data": "Four"}'))

PRINT @info;

[{"id": 1, "data": "One"}, {"id": 2, "data": "Two"}]
[{"id": 1, "data": "One"}, {"id": 2, "data": "Two"},{"id": 3, "data": "Three"},{"id": 4, "data": "Four"}]

DECLARE @json1 NVARCHAR(500)='[{"id": 1, "data": "One"}, {"id": 2, "data": "Two"}]';
DECLARE @json2 NVARCHAR(500)='[{"id": 3, "data": "Three"}, {"id": 4, "data": "Four"}]';

SELECT t.id, t.[data]
FROM
(
    SELECT * FROM OPENJSON(@json1) WITH(id int,[data] NVARCHAR(MAX)) 
    UNION ALL
    SELECT * FROM OPENJSON(@json2) WITH(id int,[data] NVARCHAR(MAX))
) t
FOR JSON PATH;

JSON_F52E2B61-18A1-11d1-B105-00805F49916B

[{"id":1,"data":"One"},{"id":2,"data":"Two"},{"id":3,"data":"Three"},{"id":4,"data":"Four"}]